The grades on a language midterm at Covington are normally distributed with $\mu = 84$ and $\sigma = 2.0$. Kevin earned a n $89$ on the exam. Find the z-score for Kevin's exam grade. Round to two decimal places.
Answer: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Kevin's exam grade by subtracting the mean $(\mu)$ from his grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{89 - {84}}{{2.0}}} $ ${ z \approx 2.50}$ The z-score is $2.50$. In other words, Kevin's score was $2.50$ standard deviations above the mean.